If the equation of the plane containing the line $$x + 2y + 3z - 4 = 0 = 2x + y - z + 5$$ and perpendicular to the plane $$\vec{r} = (\hat{i} - \hat{j}) + \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} - 2\hat{j} + 3\hat{k})$$ is $$ax + by + cz = 4$$ then $$(a - b + c)$$ is equal to

We need to find $$a - b + c$$ where $$ax + by + cz = 4$$ is the equation of the plane containing the line $$x + 2y + 3z - 4 = 0 = 2x + y - z + 5$$ and perpendicular to the plane $$\vec{r} = (\hat{i} - \hat{j}) + \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} - 2\hat{j} + 3\hat{k})$$.

Any plane through the line of intersection can be written as $$(x + 2y + 3z - 4) + \lambda(2x + y - z + 5) = 0$$.

This expands to $$(1 + 2\lambda)x + (2 + \lambda)y + (3 - \lambda)z + (-4 + 5\lambda) = 0$$.

The given parallel plane has direction vectors $$\vec{v_1} = (1, 1, 1)$$ and $$\vec{v_2} = (1, -2, 3)$$.

The normal vector is given by the cross product $$\vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{vmatrix} = (3+2, 1-3, -2-1) = (5, -2, -3)$$.

The normal to our plane is $$(1 + 2\lambda, 2 + \lambda, 3 - \lambda)$$, which must be perpendicular to $$(5, -2, -3)$$.

Hence $$5(1 + 2\lambda) - 2(2 + \lambda) - 3(3 - \lambda) = 0$$.

Expanding gives $$5 + 10\lambda - 4 - 2\lambda - 9 + 3\lambda = 0$$.

So $$11\lambda - 8 = 0 \implies \lambda = \dfrac{8}{11}$$.

Substituting back, the normal of the plane becomes $$\left(\dfrac{27}{11}, \dfrac{30}{11}, \dfrac{25}{11}\right)$$.

The constant term is $$-4 + 5 \cdot \dfrac{8}{11} = \dfrac{-44 + 40}{11} = \dfrac{-4}{11}$$.

Thus the plane equation is $$\dfrac{27}{11}x + \dfrac{30}{11}y + \dfrac{25}{11}z = \dfrac{4}{11}$$, which on multiplying by 11 yields $$27x + 30y + 25z = 4$$.

From this, $$a = 27$$, $$b = 30$$, and $$c = 25$$.

Therefore, $$a - b + c = 27 - 30 + 25 = 22$$.

The correct answer is Option B: 22.